What is area of the image of the unit disk under the mapping $(x,y)\to \left(\frac{x+y}{\sqrt{2}},\frac{y-x}{\sqrt{2}}\right)$?

Question

For $(x,y) \in \Bbb{R^2}$, let $$T(x,y)= \Big(\frac{x+y}{\sqrt2} , \frac{y-x}{\sqrt2}\Big)$$ If $\Bbb{D}$ is the unit disc in the plane , what is area of $T(\Bbb{D})$?

I was trying this question many ways, but I could not get. This question is given by my teacher as a homework. I was thinking about the area of the circle, but I don't know where I have to start and I don't have any hint to solve this question.

If anybody help me I would be very thankful to him.

Answer 1

Determinant of the matrix of $T$ it's $\frac{1}{\sqrt2}\cdot\frac{1}{\sqrt2}-\left(-\frac{1}{\sqrt2}\right)\cdot\frac{1}{\sqrt2}=1$,

which says that the area does not change.

Answer 2

This T is rotation around (0,0) for 45 degree, thus it preserves area.

Say $z=x+yi$, then $T(z) = (\cos 45+i\sin 45)\cdot z$